An inverted cylindrical cone, 52 ft deep and 26 ft across at the top, is being filled with water at a rate of 11 ft3/min. At what rate is the water rising in the tank when the depth of the water is 1 foot?
At a given time of t minutes,
let the height of water be h, let the radius of water level be r
by simple ratios,
r/h = 13/52 = 1/4
so r = h/4
given: dV/dt = 11 ft^3/min
find : dh/dt when h = 1
V = (1/3)π r^2 h
= (1/3) π (h^2/16)h
= (1/48) π h^3
dV/dt = (1/16)π h^2 dh/dt
11 = (1/16)π(1) dh/dt
dh/dt = 176/π ft/min
check my arithmetic
To find the rate at which the water is rising in the tank when the depth of the water is 1 foot, we need to calculate the rate of change of the depth with respect to time.
Let's denote the depth of the water as h (in feet) and the time as t (in minutes).
We are given that the cone is inverted, so we need to use negative values for the depth.
Given:
h = -1 ft (depth of water)
dV/dt = 11 ft^3/min (rate of change of the volume of water)
r = 26/2 = 13 ft (radius of the top of the cone)
We need to find dh/dt (the rate at which the water is rising).
The volume of a cone is given by:
V = (1/3) * π * r^2 * h
Differentiating both sides of the equation with respect to time (t), we get:
dV/dt = (1/3) * π * 2r * rh/dt
Since we want to find dh/dt, we can rearrange the equation to solve for it:
dh/dt = (3 * dV/dt)/(π * 2r * h)
Substituting the given values:
dh/dt = (3 * 11)/(π * 2 * 13 * -1) [Remember to use negative sign for depth]
Simplifying the expression:
dh/dt = -33/(26π)
Now, to get the final numerical value, we can approximate π as 3.14:
dh/dt ≈ -33/(26 * 3.14)
≈ -33/81.64
≈ -0.404 ft/min
The rate at which the water is rising when the depth is 1 foot is approximately -0.404 ft/min.
To find the rate at which the water is rising in the tank when the depth of the water is 1 foot, we can use related rates.
Let's start by visualizing the given scenario. We have an inverted cylindrical cone that is 52 ft deep and 26 ft across at the top. We need to find the rate at which the water level is rising, given that water is being filled into the cone at a rate of 11 ft³/min.
First, let's consider the information that is given about the cylinder. The cylinder is inverted, meaning the larger opening is at the bottom, and the smaller opening is at the top. We are given the depth of the water, but we need to find the rate at which the water level is rising.
To solve this problem, we can use the formula for the volume of a cone:
V = 1/3 * π * r² * h
Where:
V = Volume of the cone
π = Pi, approximately 3.14159
r = Radius of the cone's base
h = Height/Depth of the water level
Since we know the diameter of the top of the cone (26 ft across), we can find the radius by dividing it by 2:
r = 26 ft / 2 = 13 ft
Now, let's take the derivative of the formula with respect to time to find the rate at which the volume is changing with respect to time:
dV/dt = 1/3 * π * (2r * dr/dt) * h + 1/3 * π * r² * dh/dt
Where:
dV/dt = Rate of change of volume with respect to time (which is 11 ft³/min in this case since water is being filled into the cone at a rate of 11 ft³/min)
dr/dt = Rate of change of radius with respect to time (which is 0 since the radius remains constant in this problem)
h = Height/Depth of the water level (which is 1 ft in this case)
dh/dt = Rate of change of height/depth with respect to time (which is what we need to find)
Simplifying the equation and plugging in the known values:
11 ft³/min = 1/3 * π * (2 * 13 ft * 0 ft/min) * 1 ft + 1/3 * π * (13 ft)² * dh/dt
Simplifying further:
11 ft³/min = 0 + 1/3 * π * (169 ft²) * dh/dt
Now, we can solve for dh/dt, which represents the rate at which the water level is rising:
dh/dt = (11 ft³/min) / [1/3 * π * (169 ft²)]
Calculating this expression:
dh/dt = (11 ft³/min) / [(1/3) * π * 169 ft²]
= (11 ft³/min) / (0.333 * π * 169 ft²)
≈ 0.0203 ft/min
Therefore, when the depth of the water is 1 foot, the rate at which the water level is rising is approximately 0.0203 ft/min.